Transcript Physics 207: Lecture 2 Notes

Lecture 16
Goals:
• Chapter 12
 Extend the particle model to
rigid-bodies
 Understand the equilibrium of
an extended object.
 Analyze rolling motion
 Understand rotation about a
fixed axis.
 Employ “conservation of
angular momentum” concept
Assignment:
 HW7 due March 25th
 After Spring Break Tuesday:
Catch up
Physics 207: Lecture 16, Pg 1
Rotational Dynamics: A child’s toy, a physics
playground or a student’s nightmare
 A merry-go-round is spinning and we run and jump on
it. What does it do?
What principles would apply?
 We are standing on the rim and our “friends” spin it
faster. What happens to us?
 We are standing on the rim a walk towards the center.
Does anything change?
Physics 207: Lecture 16, Pg 2
Rotational Variables
 Rotation about a fixed axis:
 Consider a disk rotating about
an axis through its center:


 How do we describe the motion:
d 2


(rad/s)  vTangential /R
dt T
(Analogous to the linear case v 
dx )
dt
Physics 207: Lecture 16, Pg 3
Rotational Variables...
 Recall: At a point a distance R away from the axis of
rotation, the tangential motion:
v=R
 x= R
v=R
a=R
  constant
  0  t

R
x

2
(angular accelation in rad/s )

(angular v elocity in rad/s)
1
   0  0 t  t 2 (angular position in rad)
2
Physics 207: Lecture 16, Pg 4
Comparison to 1-D kinematics
Angular
Linear
  constant
a  constant
  0  t
  0  0 t  t
1
2
v  v0  at
2
x  x0  v0 t  12 at 2
And for a point at a distance R from the rotation axis:
x=R
v=R
aT =  R
Here aT refers to tangential acceleration
Physics 207: Lecture 16, Pg 5
System of Particles (Distributed Mass):
 Until now, we have considered the behavior of very simple
systems (one or two masses).
 But real objects have distributed mass !
 For example, consider a simple rotating disk and 2 equal
mass m plugs at distances r and 2r.

1
2
 Compare the velocities and kinetic energies at these two
points.
Physics 207: Lecture 16, Pg 9
System of Particles (Distributed Mass):
1 K= ½ m v2 = ½ m ( r)2

2 K= ½ m (2v)2 = ½ m ( 2r)2
 Twice the radius, four times the kinetic energy
K  mv  m(r )
1
2
2
1
2
2
 The rotation axis matters too!
Physics 207: Lecture 16, Pg 10
A special point for rotation
System of Particles: Center of Mass (CM)
 If an object is not held then it will rotate about the
center of mass.
 Center of mass: Where the system is balanced !
 Building a mobile is an exercise in finding
centers of mass.
m1
+
m2
m1
+
m2
mobile
Physics 207: Lecture 16, Pg 11
System of Particles: Center of Mass
 How do we describe the “position” of a system made up of
many parts ?
 Define the Center of Mass (average position):
 For a collection of N individual point like particles whose
masses and positions we know:

 mi ri
N

i 1
RCM 
M
RCM
m2
m1
r1
r2
y
x
(In this case, N = 2)
Physics 207: Lecture 16, Pg 12
Sample calculation:
 Consider the following mass distribution:
N

 mi ri

RCM  i 1
M
 XCM ˆi  YCM ˆj  ZCM kˆ
XCM = (m x 0 + 2m x 12 + m x 24 )/4m meters
RCM = (12,6)
YCM = (m x 0 + 2m x 12 + m x 0 )/4m meters
(12,12)
2m
XCM = 12 meters
YCM = 6 meters
m
(0,0)
m
(24,0)
Physics 207: Lecture 16, Pg 13
System of Particles: Center of Mass
 For a continuous solid, convert sums to an integral.
dm
y



 r dm  r dm
RCM 

M
 dm
r
x
where dm is an infinitesimal
mass element.
Physics 207: Lecture 16, Pg 14
Connection with motion...
 So for a rigid object which rotates about its center of
mass and whose CM is moving:
K TOTAL  K Rotation  K Translatio n
K TOTAL  K Rotation  MV
2
CM
1
2
For a point p rotating:

 mi ri
N

RCM  i 1
M
K R  m p v p  m p (rp )
1
2
p
2
1
2
VCM

Physics 207: Lecture 16, Pg 15
2
Rotation & Kinetic Energy
 Consider the simple rotating system shown below.
(Assume the masses are attached to the rotation axis by
massless rigid rods).
 The kinetic energy of this system will be the sum of the
kinetic energy of each piece:
4
K   mi v
1
2
i 1
2
i
 K = ½ m1v12 + ½ m2v22 + ½ m3v32 + ½ m4v42
m4

r4
m3
r3
r1
m1
r2
m2
Physics 207: Lecture 16, Pg 16
Rotation & Kinetic Energy
 Notice that v1 =  r1 , v2 =  r2 , v3 =  r3 , v4 =  r4
 So we can rewrite the summation:
4
4
K   mi v   mi r 
1
2
2
i
i 1
1
2
2 2
i
i 1
1
2
4
[  m r ]
i 1
2
i i
2
 We recognize the quantity, moment of inertia or I, and
write:
m4
K Rotational  I 
1
2
2
m3
N
I   mi ri
2

r4
r3
r1
r2
m1
m2
i 1
Physics 207: Lecture 16, Pg 17
Calculating Moment of Inertia
N
I   mi ri
2
where r is the distance from
the mass to the axis of rotation.
i 1
Example: Calculate the moment of inertia of four point masses
(m) on the corners of a square whose sides have length L,
about a perpendicular axis through the center of the square:
m
m
m
m
L
Physics 207: Lecture 16, Pg 18
Calculating Moment of Inertia...
 For a single object, I depends on the rotation axis!
 Example: I1 = 4 m R2 = 4 m (21/2 L / 2)2
I1 = 2mL2
m
m
m
m
I2 = mL2
I = 2mL2
L
Physics 207: Lecture 16, Pg 19
Moments of Inertia
 For a continuous solid object we have to add up the mr2
contribution for every infinitesimal mass element dm.
 An integral is required to find I :
dm
I   r dm
2
r
 Some examples of I for solid objects:
dr
r
L
R
Solid disk or cylinder of mass M
and radius R, about
perpendicular axis through its
center.
I = ½ M R2
Use the table…
Physics 207: Lecture 16, Pg 22
Exercise Rotational Kinetic Energy
 We have two balls of the same mass. Ball 1 is attached
to a 0.1 m long rope. It spins around at 2 revolutions per
second. Ball 2 is on a 0.2 m long rope. It spins around at
2 revolutions per second.
2
K  12 I 
 What is the ratio of the kinetic energy
of Ball 2 to that of Ball 1 ?
A. ¼
B. ½
C. 1
D. 2
Ball 1
E. 4
I   mi ri
2
i
Ball 2
Physics 207: Lecture 16, Pg 24
Exercise Rotational Kinetic Energy
 K2/K1 = ½ m r22 / ½ m r12 = 0.22 / 0.12 = 4
 What is the ratio of the kinetic energy of Ball 2 to
that of Ball 1 ?
(A) 1/4 (B) 1/2
Ball 1
(C) 1
(D) 2
(E) 4
Ball 2
Physics 207: Lecture 16, Pg 25
Exercise
Work & Energy
 Strings are wrapped around the circumference of two solid disks
and pulled with identical forces, F, for the same linear distance,
d.
Disk 1 has a bigger radius, but both are identical material (i.e.
their density r = M / V is the same). Both disks rotate freely
around axes though their centers, and start at rest.
 Which disk has the biggest angular velocity after the drop?
W  F d = ½ I 2
2
1
(A) Disk 1
(B) Disk 2
(C) Same
start
finish
F
F
d
Physics 207: Lecture 16, Pg 28
Exercise
Work & Energy
 Strings are wrapped around the circumference of two solid
disks and pulled with identical forces for the same linear
distance.
Disk 1 has a bigger radius, but both are identical material (i.e.
their density r = M/V is the same). Both disks rotate freely
around axes though their centers, and start at rest.
 Which disk has the biggest angular velocity after the drop?
W = F d = ½ I1 12 = ½ I2 22
2
1
1 = (I2 / I1)½ 2 and I2 < I1
(A) Disk 1
(B) Disk 2
(C) Same
start
finish
F
F
d
Physics 207: Lecture 16, Pg 29
Lecture 16
K TOTAL  K Rotational  K Translatio nal
K TOTAL  K Rotational  MV
2
CM
1
2
K Rotational  I 
1
2
Assignment:
 HW7 due March 25th
 For the next Tuesday:
Catch up
2
I   mi ri
2
i
Physics 207: Lecture 16, Pg 30
Lecture 16
Assignment:
 HW7 due March 25th
 After Spring Break Tuesday: Catch up
Physics 207: Lecture 16, Pg 31